Real object Virtual image Divergent Lens 1 p 1 q 1 f m h h q p Diverging Lens f

Real object virtual image divergent lens 1 p 1 q 1 f

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Real object Virtual image Divergent Lens 1 p + 1 q = 1 f m = h h = - q p Diverging Lens 0 > f > p > 0 - f < q < 0 0 < m < 1 0 > p > - f 0 < q < 1 < m < - f > p > -∞ -∞ < q < - f < m < 0 012 (part 2 of 3) 10.0 points The image distance is 1. - 17 27 f . 2. - 18 25 f . 3. - 16 23 f . 4. - 12 17 f . 5. - 9 14 f . 6. - 14 23 f . 7. - 17 23 f . 8. - 11 18 f . correct 9. - 13 18 f . 10. - 17 26 f . Explanation:
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Munoz (gm7794) – HW10 – TSOI – (92515) 6 0 11 7 f f f p q - 11 18 f Basic Concepts: 1 p + 1 q = 1 f , where f < 0 for a divergent lens. Solution: 1 q 1 = - 1 p 1 - 1 f = - (7) 11 f - 1 f = - (11) - (7) 11 f =-1811fq1=-1118f .013(part 3 of 3) 10.0 pointsThe magnification is1.517.2.1027.3.825.4.27.5.413.6.720.A convergent lens which has a focal point f.An object is placed at distancep=32ftothe left of the lens.pfThe image distanceqcan quickly be shownto be 3f .Now place a convergent lens with a samefocal lengthfat a distanced=fbehind thefirst lens.dpf#1#2 7. 7 18 . correct 8. 7 19 . 9. 11 30 . 10. 10 29 . Explanation: The magnification m of this lens is m = - q 1 p 1 = - 11 - 18 f 11 7 f = - 7 - 18 = 7 18 Note: This problem uses only real objects. keywords: 014 10.0 points
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Munoz (gm7794) – HW10 – TSOI – (92515) 7 Determine q 2 ; i.e. , the image location mea- sured with respect to the second lens #2. 1. q 2 = 1 2 f 2. q 2 = 5 9 f 3. q 2 = 7 9 f 4. q 2 = 4 9 f 5. q 2 = 1 3 f 6. q 2 = 2 3 f correct 7. q 2 = 8 9 f 8. q 2 = 1 9 f 9. q 2 = f 10. q 2 = 2 9 f Explanation: p 2 = - ( q 1 - d ) , virtual = - (3 f - f ) = - 2 f , therefore 1 p 2 + 1 q 2 = 1 f 1 q 2 = 1 - p 2 + 1 f = 1 2 f + 1 f = 3 2 f q 2 = 2 3 f . keywords: 015 10.0 points Consider the light rays depicted in the figure. What type of reflecting or refracting surface is depicted? 1. converging mirror 2. Unable to determine. 3. diverging lens 4. diverging mirror correct 5. plane mirror 6. converging lens Explanation: The parallel light rays reflect and and di- verge as if originating from a single point, which must be the focal point of a diverging (convex) mirror. 016 10.0 points An ordinary magnifying glass produces an image which is 1. virtual and erect correct 2. virtual and inverted 3. real and erect 4. none of these 5. real and inverted Explanation: A magnifying glass is a double convex lens. The object is located just inside the focal length. Thus the image is virtual, erect, and
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Munoz (gm7794) – HW10 – TSOI – (92515)8the magnification is larger than 1.
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  • Spring '08
  • morrison
  • Physics, Munoz

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